# What consistency results follow the assumption: $\forall\alpha(\aleph_{\alpha+1}\nleq2^{\aleph_\alpha})$?

In a recent question on Math.SE it was asked whether or not For every infinite cardinal $\mathfrak m$ there is no $\aleph$ number, $\kappa$, such that $\mathfrak m<\kappa<2^{\mathfrak m}$.

By requiring that $\mathfrak m<\kappa$ we invariably say that $\mathfrak m$ is an $\aleph$, and so this translates to the following statement,

$$\forall\alpha(\aleph_{\alpha+1}\nleq2^{\aleph_\alpha})\qquad(\bullet)$$

As it turns out this is indeed possible. Starting from a $3$-huge (and remarking this can be improved to an almost huge), Apter showed that it is consistent that every successor cardinal is Ramsey, and from this we can show that indeed $(\bullet)$ holds while the axiom of choice fail.

But what about the inverse direction? Well, we can show that every successor cardinal is in fact a limit cardinal in any inner model of $\sf AC$. In particular $V$ and $L$ disagree on the successors of singular cardinals, so $0^\#$ exists. If there are regular cardinals, then they are strongly inaccessible in any inner model satisfying $\sf AC$.

What sort of large cardinals can we draw from $(\bullet)$? Is there something we can say about regular cardinals of $V$, in addition to them being strongly inaccessible in inner models of $\sf AC$?

As you mention, the hypothesis ($\bullet$) implies that for every singular cardinal $\kappa$ and every inner model $W$ of $\mathsf{ZFC}$, we have $\kappa^{+W} < \kappa^+$. From this consequence, I suspect that it might follow by the argument of Busche and Schindler [1] that there is a proper class model of $\mathsf{AD}$ in a forcing extension of $V$.

The reason I say this is that in the paper [1, Subsections 2.2.2 and 2.2.3] it seems like this consequence might suffice for their argument when it is applied to the $\mathsf{ZFC}$ model $W = \text{Lp}(A_0)$ for a sufficiently closed singular cardinal $\kappa$ and a sufficiently complicated set of ordinals $A_0 \subset \kappa$ that they consider. The notation $\text{Lp}(A_0)$ refers to the "lower part mouse" over $A_0$, which is the relevant generalization of $L$ in the argument for $0^\sharp$ that you mention.

I haven't read the paper carefully, so if you want to know whether this is true then you should ask them.

The next natural goal for a consistency strength lower bound would be $\mathsf{AD}_\mathbb{R}$. I don't think anyone knows exactly how to get this from ($\bullet$) at the moment; the absence of choice may limit the available methods somewhat. It might be a good problem. (Again the relevant consequence would probably be that inner models of choice do not compute successors of singulars correctly.)

In terms of large cardinals, the theory $\mathsf{ZF} + \mathsf{AD}$ is equiconsistent with $\mathsf{ZFC} + {}$"there are infinitely many Woodin cardinals," and $\mathsf{ZF} + \mathsf{AD}_\mathbb{R}$ is equiconsistent with $\mathsf{ZFC} + {}$"there is a cardinal $\lambda$ that is a limit of Woodin cardinals and $\mathord{<}\lambda$-strong cardinals." However, current methods typically proceed in terms of building determinacy models rather than building models of $\mathsf{ZFC} + {}$large cardinals directly.

It may be interesting to note that ($\bullet$) implies $\neg\square_\kappa$ for every infinite cardinal $\kappa$ (so in particular we get failures of square at singular cardinals.) The reason is that from a square sequence we can recursively define a sequence of surjections $f_\alpha :\kappa \to \alpha$ for $\alpha < \kappa^+$, using our clubs at limit stages to piece together the surjections that we have already constructed into a new surjection. (Coherence is superfluous.) Then we can define an injection from $\kappa^+$ to $\mathcal{P}(\kappa \times \kappa)$, or equivalently to $\mathcal{P}(\kappa)$, by $\alpha \mapsto \{(\gamma,\delta) \in \kappa \times \kappa : f_\alpha(\gamma) < f_\alpha(\delta)\}$.

[1] Busche, Daniel, and Ralf Schindler. The strength of choiceless patterns of singular and weakly compact cardinals, Annals of Pure and Applied Logic, 159(1), 198–248, 2009

• Awesome. Is $(\bullet)$ somewhat a known axiom, or can you just say these things because of some obvious consequences like failure of the covering lemma for all sort of inner models? – Asaf Karagila May 9 '15 at 7:09
• ($\bullet$) is known in certain contexts (typically below $\Theta$) as the boldface $\mathsf{GCH}$. But I don't know if anyone has looked at its consistency strength specifically. So what I said was just based on its consequences for failure of covering. – Trevor Wilson May 9 '15 at 7:18
• Thanks, that's interesting. Do you have some educated guess to make (mine is that $(\bullet)$ is roughly equiconsistent with a proper class of Woodin cardinals). – Asaf Karagila May 9 '15 at 7:29
• @YairHayut I think it is weaker without $\mathsf{AC}$. Woodin has proved that $\mathsf{AD} + \mathsf{DC} + {}$"$\omega_1$ is supercompact" is consistent relative to a proper class of Woodin limits of Woodins. – Trevor Wilson May 9 '15 at 17:51
• It's interesting to note that if the HOD conjecture is true, then this result is inconsistent with the existence of an extendible cardinal, as it implies $V$ is far far away from $\sf HOD$. – Asaf Karagila Nov 1 '15 at 17:55